# Poles, Polars and Quadrilaterals

Let there be a quadrilateral ABCD inscribed in a circle of radius R centered at O. Draw the tangents to the circle at the vertices of the quadrilateral and assume they meet successively at points A', B', C' and D' as in the applet below (have the box "Show names" checked.)

What if applet does not run? |

A'B'C'D' is a quadrilateral circumscribed around the given circle. (The right term for this is *inscriptible*: A polygon (a quadrilateral, in particular) is inscriptible if there exists a circle tangent to all the side lines of the equilateral.)

The two quadrilaterals stand to each other in a very definite relationship, viz., sides of each of them are the polars of the vertices of the other, so that too, vertices of one are the poles of the side lines of the other. More accurately, the relationships are summarized in the table below.

Pole | Polar | Why | ||
---|---|---|---|---|

A | D'A' | The polar of a point on the circle is tangent to the circle at that point. | ||

B | A'B' | |||

C | B'C' | |||

D | C'D' | |||

A' | AB | For a point outside the circle, the polar joins the tangency points of the tangents from the point to the circle. | ||

B' | BC | |||

C' | CD | |||

D' | DA |

There's more. Define P as the intersection of the diagonals AC and BD, M the intersection of AB and CD (which may be at infinity if the latter are parallel), and K the intersection of AD and BC.

What if applet does not run? |

Pole | Polar | Why | ||
---|---|---|---|---|

M | KP | From a straightedge construction of the polar. | ||

K | MP | |||

P | KM | The polar of the intersection of two polars is the line through their poles. |

Now, since the lines BC, KP, AD, KM all pass through K, their poles B', M, D', and P are collinear. Similarly, as the poles of AB, CD, MP, and KM, points A', C', K, and P are also collinear. As a consequence, the diagonals of the circumscribed quadrilateral A'B'C'D' meet at the same point as those of the inscribed quadrilateral ABCD. (We have this fact also proven in a different way.)

Let L and N be the poles of AC and BD, respectively. Since P lies at the intersection of AC and BD, L and N lie on the polar of P, which is KM. The four points K, L, M, and N are therefore collinear.

Consider now the complete quadrilateral formed by the side lines of the circumscribed quadrilateral A'B'C'D'. LN is one of its diagonals. The other two, A'C' and B'D', meet the latter at points K and M, respectively. We thus have that L and N are harmonic conjugate with respect to K and M:

(1)

KL/KN : ML/MN = -1.

Given the multitude of lines included in the above configuration, there are in fact several complete quadrilaterals to which the same consideration applies. It follows that the configuration contains several harmonically conjugate pairs of points. Among others, there are two involving points on the diagonals of A'B'C'D':

(2)

KC'/KA' : PC'/PA' = -1, MB'/MD' : PB'/PD' = -1.

## Poles and Polars

- Poles and Polars
- Brianchon's Theorem
- Complete Quadrilateral
- Harmonic Ratio
- Harmonic Ratio in Complex Domain
- Inversion
- Joachimsthal's Notations
- La Hire's Theorem
- La Hire's Theorem, a Variant
- La Hire's Theorem in Ellipse
- Nobbs' Points, Gergonne Line
- Polar Circle
- Pole and Polar with Respect to a Triangle
- Poles, Polars and Quadrilaterals
- Straight Edge Only Construction of Polar
- Tangents and Diagonals in Cyclic Quadrilateral
- Secant, Tangents and Orthogonality
- Poles, Polars and Orthogonal Circles
- Seven Problems in Equilateral Triangle, Solution to Problem 1

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